一、leetcode刷题：

1.leetcode题目 121.买卖股票的最佳时机 . - 力扣（LeetCode）(easy）

解决：

class Solution:def maxProfit(self, prices: List[int]) -> int:cur_min = prices[0]maxx = 0for i in range(1,len(prices)):maxx = max(maxx,prices[i]-cur_min)cur_min = min(cur_min,prices[i])return maxx

2.leetcode题目 122.买卖股票的最佳时机II . - 力扣（LeetCode）(medium）

解决：

贪心算法：

class Solution:def maxProfit(self, prices: List[int]) -> int:summ = 0cur_min = prices[0]for i in range(1,len(prices)):cur_min = min(cur_min,prices[i])if prices[i] > cur_min:summ += prices[i] - cur_mincur_min = prices[i]return summ

动态规划：

class Solution:def maxProfit(self, prices: List[int]) -> int:dpf = [0]*len(prices)  ##当天无持仓时的最大资产dpg = [0]*len(prices)  ##当天持仓时的最大资产dpg[0] = -prices[0]for i in range(1,len(prices)):dpf[i] = max(dpf[i-1],dpg[i-1] + prices[i])dpg[i] = max(dpg[i-1],dpf[i-1] - prices[i])return dpf[len(prices)-1]

3.leetcdoe题目 123.买卖股票的最佳时机III （medium):

动态规划：

class Solution:def maxProfit(self, prices: List[int]) -> int:n = len(prices)buy1 = -prices[0]buy2 = -prices[0]sell1=0sell2 = 0for p in prices:buy1 = max(buy1,-p)sell1 = max(sell1,buy1 + p)buy2 = max(buy2,sell1-p)sell2 = max(sell2,buy2+p)return sell2